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### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 13: Limits from equations (factoring & rationalizing)

# Finding limits by factoring (cubic)

Sal finds the limit of (x³-1)/(x²-1) at x=1 by factoring and simplifying the expression. Created by Sal Khan.

## Want to join the conversation?

• Where can I watch a tutorial on the kind of division Sal does at ?
• is x^3 - y^3 = (x-y)(x² + xy + y²) ?
• Yes, if you need proof:
(x-y)^3=x^3 - y^3 - 3xy(x-y)
=> (x - y)^3 + 3xy(x-y) = x^3 - y^3
=> x^3 - y^3 = (x - y) [(x - y)^2 + 3xy]
=> x^3 - y^3 = (x - y)(x^2 + y^2 - 2xy + 3xy)
=> x^3 - y^3 = (x - y)(x^2 + xy + y^2 )
• At , Sal writes when x =/= 1. But after simplification, even x=-1 will give an undefined answer. Why do we not take that into consideration? How do we know what to put conditions for and what not to put conditions for.
I understand that since the question deals with limit of x going to 1, -1 is not that important, but my question is still how do we generally know what to put conditions for.
• He only stated the restriction for the factor that cancelled. Since it cancelled it is no longer obvious so you must state it explicitly.

Since the x+1 factor is still there, the x=/= -1 restriction is obvious by looking at the problem.
• how do I do the problem:
Find the limit:
lim as x approaches 0: {1/(3+x)}-(1/3)/x

fractions in the numerator...I forget this algebra/simplifying
• The problem you have is that you don't want to divide by 0, right? So lets try to get rid of the x'es in the numerator, because if we do that we wont have any problem with letting x go to 0 am I right?
So my suggestion would be multiplying both denominator and numerator with x and see what happens! :)
• I read a popular anomaly in maths where the end result is 2=1 when we take A=B.
It goes like this- Let A=B
=> A^2 =AB => A^2 - B^2 = AB - B^2 => (A+B)(A-B) = B (A-B) ------ Step n
=> A+B=B
=> 2B=B => 2=1

The explanation given there was that in step n we divided both sides by zero. As A-B=0, therefore this anomaly came up.

Here too aren't we doing the same thing?
• Good question. At , Sal is careful to point out that the division he is performing is valid only if we assume x does not equal 1. We make that assumption when performing a limit calculation because we aren't trying to find the value when x = 1. Instead, we're trying to find the value that is approached as x gets closer and closer to 1. As long as we allow for some difference between x and 1, we aren't dividing by zero and the operation is valid.

Now it may seem like a contradiction that Sal goes on after that to compute the limit by plugging 1 into the resulting fraction, but again you have to keep in mind that he isn't calculating the value of the fraction at 1, he's calculating the value of the limit at 1, which means finding the value that the surrounding points gravitate towards. Plugging in 1 at this point gives us that value.
• Why was he allowed to use 1 if he said x cannot equal one?
• He was allowed to substitute 1 for x even though there was a restriction of "x does not equal 1" because he was finding the limit of the original expression as x approaches 1. To find the limit, he can substitute 1 for x into the new expression for which x=1 is defined to get the limit for the old expression for which x=1 is not defined. However, this is not the value of the old expression at x=1: It is the limit. That's why he can substitute 1 for x into the new expression despite the restriction.
• Would anyone able to show me why the following limit is 9? I would be much appreciate if steps are also shown. Thanks!!
Lim (sin ^2 (3x)) / x^2
x -> 0
• (In what follows, it is assumed that `x → 0` in every limit statement, if not otherwise stated.)

One way to prove the identity is by making use of the following preliminary relations:
`1. lim ( sin(x) / x ) = 1;`
`2.` If `lim ƒ(x) = F` and `lim g(x) = G`, both as `x → a`, then `lim ƒ(x)g(x) = FG` as `x → a`, where `a` is any real number.

The fact that `lim ( sin²(3x) / x² ) = 9` may now be deduced by rewriting `sin²(3x) / x²` to a form we recognise.

` I)` Properties `1.` and `2.` imply that `lim ( sin²(x) /`` x² ) = 1`. To see this, observe that `lim sin²(x) / x² = lim ( sin(x) / x )² = lim ( sin(x) / x ) · ( sin(x) / x )`. By properties `1.` and `2.`, we have, then, `lim sin²(x) / x² = (lim sin(x) / x) · (lim sin(x) / x) = 1 · 1 = 1`.

` II)` We now compute `lim sin²(3x) / (3x)²`. Let `u = 3x`. Then `u → 0` as `x → 0`; hence `lim sin²(3x) / (3x)² = lim sin²(u) / u² = 1`, which follows from `I)`.

`III)` Observe that `sin²(3x) / x² = sin²(3x) / ([1/9] · 9x²) = 9 · ( sin²(3x) / (3x)² )`. By properties `II)` and `2.`, it follows that `lim sin²(3x) / x²` exists and equals `(lim 9) · (lim sin²(3x) / (3x)²) = 9 · 1 = 9`.

This completes the proof.
• If we differentiate the numerator and denominator by L'hospital rule that would give the same answer right?
• If you are referring to the problem in the video, that is a 0/0 form and thus is valid for l'Hopital's. So, yes, you could do it that way (and I would definitely solve it that way)

lim x→1 (x³-1)/(x²-1)
= lim x→1 (3x²)/(2x)
= ³⁄₂
However, you should not discount the techniques Sal used in the video because you cannot use l'Hopital's on many difficult limits, so it is good to know the techniques for finding the limits without l'Hopital's.

Furthermore, even when you can use l'Hopital's, it is not always the easiest method.
• at how does x^2-1 becomes (x-1)(x+1)?
(1 vote)
• I'm sorry, but if you have not yet had enough algebra to know how to factor a difference of squares, then you need to go back and study that material before proceeding to study calculus.

Studying calculus requires that you have already mastered at the very least Algebra I and II and basic Geometry. Without mastering those, you will not be able to understand the content in a Calculus course. Preferably, you would also have had a Trigonometry / Pre-calculus course.

Calculus requires mastery of the ability to factor polynomials and rational expressions, so it will not explain how to do this. Kahn Academy has videos on factoring quadratic expressions that you should review.